Japan 2013 Initial Round

Using the coefficients $a_0,a_1,a_2,\dots ,a_9$, we can write $$(x+1{)}^3(x+2{)}^3(x+3{)}^3=a_0+a_{1}x+a_2{x}^2+\cdots +a_8{x}^8+a_9{x}^9.$$ Substitution of $x=1$ and $x=-1$ in the equation above yields, respectively $$a_0+a_1+a_2+\cdots +a_8+a_9=2^3\cdot 3^3\cdot 4^3$$ $$a_0-a_1+a_2-a_3+a_4-a_5+a_6-a_7+a_8-a_9=0.$$ From these identities we obtain $a_0+a_2+a_4+a_6+a_8=\frac{1}{2}(2^3\cdot 3^3\cdot 4^3+0)=6912$. By substituting $x=0$ in the equation above, we also get $a_0=1^3\cdot 2^3\cdot 3^3=216$. We can therefore conclude that $a_2+a_4+a_6+a_8=6912-216=6696$.